/*
给你一个嵌套的整型列表。请你设计一个迭代器，使其能够遍历这个整型列表中的所有整数。

列表中的每一项或者为一个整数，或者是另一个列表。其中列表的元素也可能是整数或是其他列表。

 

示例 1:

输入: [[1,1],2,[1,1]]
输出: [1,1,2,1,1]
解释: 通过重复调用 next 直到 hasNext 返回 false，next 返回的元素的顺序应该是: [1,1,2,1,1]。
示例 2:

输入: [1,[4,[6]]]
输出: [1,4,6]
解释: 通过重复调用 next 直到 hasNext 返回 false，next 返回的元素的顺序应该是: [1,4,6]。

*/

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 */
class NestedInteger {
public:
    // Return true if this NestedInteger holds a single integer, rather than a nested list.
    bool isInteger() const;
    // Return the single integer that this NestedInteger holds, if it holds a single integer
    // The result is undefined if this NestedInteger holds a nested list
    int getInteger() const;
    // Return the nested list that this NestedInteger holds, if it holds a nested list
    // The result is undefined if this NestedInteger holds a single integer
    const vector<NestedInteger> &getList() const;
};

#include "stdc++.h"

/* dfs
*/
class NestedIterator {
public:
    NestedIterator(vector<NestedInteger> &nestedList) {
        recursionBuildNums(nestedList); // dfs
    }

    int next() {
        if (hasNext()) {
            return nums[index++];
        } else {
            return -1;
        }
    }

    bool hasNext() {
        return index != length;
    }
private:
    vector<int> nums;
    int index{0};
    int length{0};

    void recursionBuildNums(const vector<NestedInteger> &nestedList) {
        for (const NestedInteger ni: nestedList) {
            if (ni.isInteger()) {
                nums.push_back(ni.getInteger());
                ++length;
            } else {
                recursionBuildNums(ni.getList());
            }
        }
    }
};

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i(nestedList);
 * while (i.hasNext()) cout << i.next();
 */